In my calculus book has been written: "For finding limits sometimes we can change $$\sqrt[m]{a_m x^m + a_{m - 1} x^{m-1} + a_{m - 2} x^{m-2} + \cdots + a_0} $$ to $\sqrt[m]{a_m}\left(x + \dfrac{a_{m - 1}}{ma_m}\right)$ for odd $m$ and $\sqrt[m]{a_m}\left|x + \dfrac{a_{m - 1}}{ma_m}\right|$ for even $m$."
First of all I want to know how these formulas were built. After that , when I can use them without any mistake in value of limit. I searched in the internet but I didn't find these formulas.
I would strongly advise you against adding such cheap tricks to your bag of tools. It is very unfortunate (for students) that most introductory calculus textbooks tend to provide such non-sense without any justification. I had a lot of struggle with such books in India when I was studying calculus (and I guess your book is also by an Indian author). Almost always such tricks don't provide all the context and the exact rigorous formulation (and the proof).
I will state the desired formulas along with proofs below. In what follows $n$ is a positive integer.
Let $P(x)$ be a polynomial of degree $n$ and assume that $n$ is odd. Also let $a, b$ be the coefficients of $x^{n}$ and $x^{n - 1}$ in $P(x)$ respectively. We thus have $$P(x) = ax^{n} + bx^{n - 1} + Q(x)\tag{1}$$ where $Q(x)$ is a polynomial of degree less than $n - 1$. Then we have the following limit formula: $$\lim_{x \to \infty}\sqrt[n]{P(x)} - a^{1/n}x = \frac{b}{n}a^{1/n - 1}\tag{2}$$ The proof can be given by putting $t = 1/x$ so that $t \to 0^{+}$ as $x \to \infty$ and hence the desired limit is given by $$\lim_{t \to 0^{+}}\frac{\sqrt[n]{a + bt + t^{n}Q(1/t)} - a^{1/n}}{t}$$ Now we can note that $t^{n}Q(1/t)$ is a polynomial of degree $n$ or less which has $t^{2}$ as a factor. So we can write $t^{n}Q(1/t) = t^{2}R(t)$ for some polynomial $R(t)$ of degree less than $n - 1$. Next we put $u = a + bt + t^{2}R(t)$ so that $u \to a$ as $t \to 0^{+}$. We can now see that $(u - a)/t \to b$ and thus the desired limit is reduced to $$\lim_{u \to a}\frac{u^{1/n} - a^{1/n}}{u - a}\cdot\frac{u - a}{t}$$ and this is clearly equal to $b\cdot(1/n)a^{1/n - 1}$ as expected. Here we have used the standard limit formula $$\lim_{x \to a}\frac{x^{r} - a^{r}}{x - a} = ra^{r - 1}\tag{3}$$ which is valid for all values of $r$ and it is this standard limit formula which you should put in your bag of tools.
Some people try to think of equation $(2)$ as meaning $$\sqrt[n]{P(x)} - a^{1/n}x \approx \frac{b}{n}a^{1/n - 1}$$ or $$\sqrt[n]{P(x)} \approx a^{1/n}\left(x + \frac{b}{na}\right)\tag{4}$$ All this is fine upto this point, but a blunder is committed when we say that we can replace LHS of $(4)$ by RHS of $(4)$ while evaluating certain limits. Remember that in mathematics you can replace a thing $A$ by another thing $B$ only when $A = B$ and not otherwise. The equation $(4)$ does not describe an equality between two things rather it says that one thing is an approximation of another under certain circumstances. It is however perfectly valid to replace the LHS of $(2)$ i.e. $\lim_{x \to \infty}\sqrt[n]{P(x)} - a^{1/n}x$ by RHS of $(2)$ i.e $(b/n)a^{1/n - 1}$ because these are equal via equation $(2)$.
You should try to formulate and prove the assertion for the case when $n$ is an even positive integer. In this case the leading coefficient of $P(x)$ must be positive.